Let V be a finite-dimensional vector space over R. The dual of V = V^{∗} = {f : V → R | f linear } Note V^{∗} is a vector space. The elements of V^{∗} are called covectors. If e_{1}, ..., e_{n} basis for V, then w_{1}, ..., w_{n} basis for V^{∗} where w_{i} : V → R where

w_{i}(e_{j}) = δ_{ij} = \{ \begin{matrix}1 & i = j \\ 0 & i ≠ j \end{matrix}

Let F_{∗} : V → W be a linear map between vector spaces The dual map map is F^{∗} : W^{∗} → V^{∗}, F (g) = g ∘ F. F_{∗} is injective implies F^{∗} injective F_{∗} is surjective implies F^{∗} surjective (G_{∗} ∘ F_{∗})^{∗} = F^{∗} ∘ G^{∗}. d : V → (V^{∗})^{∗}, d(v) = h where h : V^{∗} → R, h(f) = f(v). Thus (V^{∗})^{∗} is naturally isomorphic to V. Defn: The dual of T_{p}M = T_{p}^{∗}M is the cotangent space to M at p. If \frac{∂}{∂x_{1}} , ..., \frac{∂}{∂x_{m}} is a basis for T_{p}M, then the dual basis will be denoted dx_{1}, ..., dx_{m}.

B is bilinear if B(cv_{1} + dv_{2}, w) = cB(v_{1}, w) + dB(v_{2}, w) B(v, cw_{1} + dw_{2}) = cB(v, w_{1}) + dB(v, w_{2}) Thus B((v_{1}, w_{1}) + (v_{2}, w_{2})) = B(v_{1} + v_{2}, w_{1} + w_{2}) = B(v_{1}, w_{1} + w_{2}) + B(v_{2}, w_{1} + w_{2}) = B(v_{1}, w_{1}) + B(v_{1}, w_{2}) + B(v_{2}, w_{1}) + B(v_{2}, w_{2}) B is linear if B((v_{1}, w_{1}) + (v_{2}, w_{2})) = B((v_{1}, w_{1})) + B((v_{2}, w_{2})) B(c(v_{1}, w_{1})) = cB((v_{1}, w_{1}) Let (V × W)^{b} = {B : V × W → R | B bilinear } V ⊗ W = [(V × W)^{b}]^{∗} = {h : (V × W)^{b} → R | h linear } If v ∈ V, w ∈ W, let v ⊗ w : (V × W)^{b} → R, (v ⊗ w)(B) = B(v, w) Prop: v ⊗ w ∈ V ⊗ W. Proof: (v ⊗ w)(c_{1}B_{1} + c_{2}B_{2}) = (c_{1}B_{1} + c_{2}B_{2})(v, w) = c_{1}B_{1}(v, w) + c_{2}B_{2}(v, w) = c_{1}(v ⊗ w)(B_{1}) + c_{2}(v ⊗ w)(B_{2}). Prop: Σ_{i=1}^{n}r_{i}(v_{i} ⊗ w_{i}) ∈ V ⊗ W. Proof: (Σ_{i=1}^{n}r_{i}(v_{i} ⊗ w_{i}))(c_{1}B_{1} + c_{2}B_{2}) = Σ_{i=1}^{n}r_{i}(v_{i} ⊗ w_{i})(c_{1}B_{1} + c_{2}B_{2}) = Σ_{i=1}^{n}c_{1}r_{i}(v_{i} ⊗w_{i})(B_{1})+c_{2}r_{i}(v_{i} ⊗w_{i})(B_{2}) = c_{1}(Σ_{i=1}^{n}r_{i}(v_{i} ⊗w_{i}))(B_{1})+c_{2}(Σ_{i=1}^{n}r_{i}(v_{i} ⊗ w_{i}))(B_{2}). Note (Σ_{i=1}^{n}r_{i}(v_{i} ⊗ w_{i}))(B) = Σ_{i=1}^{n}r_{i}(v_{i} ⊗ w_{i})(B) = Σ_{i=1}^{n}r_{i}B(v_{i}, w_{i}) Prop: The product ⊗ is bilinear: Proof: ((c_{1}v_{1} + c_{2}v_{2}) ⊗ w)(B) = B(c_{1}v_{1} + c_{2}v_{2}, w) = c_{1}B(v_{1}, w) + c_{2}B(v_{2}, w) and ... Claim: V ⊗ W =< v ⊗ w | v ∈ V, w ∈ W,

(c_{1}v_{1} + c_{2}v_{2}) ⊗ w = c_{1}(v_{1} ⊗ w) + c_{2}(v_{2} ⊗ w), v ⊗ (c_{1}w_{1} + c_{2}w_{2}) = c_{1}(v ⊗ w_{1}) + c_{2}(v ⊗ w_{2}) >

=< v ⊗ w | v ∈ V, w ∈ W, (v_{1} + v_{2}) ⊗ w = v_{1} ⊗ w + v_{2} ⊗ w, v ⊗ (w_{1} + w_{2}) = v ⊗ w_{1} + v ⊗ w_{2}, (cv) ⊗ w = c(v ⊗ w) = v ⊗ (cw) > Let v_{1}, ..., v_{m} be a basis for V, w_{1}, ..., w_{n} basis for W. If B is a bilinear form, then B is uniquely determined by the nm values B(v_{i}, w_{j}). Thus dim(V × W)^{b} = nm = V ⊗ W. Hence a basis for V ⊗ W is {v_{i} ⊗ w_{j} | i = 1, ..., m, j = 1, ..., n} Thus V ⊗ W = {Σc_{ij}v_{i} ⊗ w_{j} | (v_{1} + v_{2}) ⊗ w = v_{1} ⊗ w + v_{2} ⊗ w,

v ⊗ (w_{1} + w_{2}) = v ⊗ w_{1} + v ⊗ w_{2}, (cv) ⊗ w = c(v ⊗ w) = v ⊗ (cw)}

The universal mapping property: There exists ϕ, V ⊗ W such that ϕ is bilinear and given any bilinear map: B : V × W → U, there exists a map A : V ⊗ W → U such that A ∘ ϕ = B. Moreover ϕ and V ⊗ W are unique in the sense that if X and ψ satisfy the universal mapping property, then there exists an isomorphism f : V ⊗ W → X such that f ∘ ϕ = ψ Fix B : V × W → U and take f : U → R ∈ U^{∗} Then f ∘ B : V × W → R is bilinear: (f ∘ B)(c_{1}v_{1} + c_{2}v_{2}, w) = f(c_{1}B(v_{1}, w) + c_{2}B(v_{2}, w)) = c_{1}(f ∘ B)(v_{1}, w) + c_{2}(f ∘ B)(v_{2}, w) Similarly (f ∘ B)(v, c_{1}w_{1} + c_{2}w_{2}) = c_{1}(f ∘ B)(v, w_{1}) + c_{2}(f ∘ B_{(}v, w_{2}) Thus f ∘ B ∈ (V × W)^{b} α : U^{∗} → (V × W)^{b}, α(f) = f ∘ B Thus β : V ⊗ W → (U^{∗})^{∗}, β The tensor algebra = T (V) = ⊕_{k=0}^{∞}(⊗^{k}V) c_{0} + c_{1}v_{i} + Σc_{ij}v_{i} ⊗ v_{j} + ... + Σc_{i_{1}...i_{p}}v_{i_{1}} ⊗ ... ⊗ v_{i_{p}} (v_{1} ⊗ ... ⊗ v_{p})(u_{1} ⊗ ... ⊗ u_{q}) = v_{1} ⊗ ... ⊗ v_{p} ⊗ u_{1} ⊗ ... ⊗ u_{q} A subring I of a ring R is an ideal of R if ar ∈ I and ra ∈ I for all a ∈ I, r ∈ R. Let I(V) = ideal generated by {v ⊗ v | v ∈ V } The exterior algebra of V is the quotient Λ^{∗}V = T (V)/I(V)